2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1

Students can Download Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1

2nd PUC Maths Linear Programming NCERT Text Book Questions and Answers Ex 12.1

Solve the following Linear Programming Problems graphically:

Question 1.
Maximize Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥0.
Answer:
x + y = 4

X231
y213

put x = 0 y = 0
since 0 + 0 ≤ 0, equation is true.
∴ This will be shaded towards 0.
At A (4,0), Z = 3 x 4 = 12
At B (0,4), Z = 4 x 4= 16
∴  Z is maximised at B (0,4) = 16
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.1

a

KSEEB Solutions

Question 2.
Minimize Z = -3x + 4y
subject to x + 2y ≤ 8; 3x + 2y ≤ 12; x ≥ 0, y ≥ 0.
Answer:
x + 2y = 8

X240
Y324

0 + 0 < 8
true towards 0

3x +  2y = 12

X240
y906

0 + 0 <12
true shaded towards 0
ABCD is the solution region
x + 2y = 8
Put x = 0 ⇒ A(-0,4)
Put y = 0 ⇒ B(8,0)
At A (0,4) ⇒ Z= 16
At B (2, 3) ⇒ Z =-3 x 2+ 4 x 3 = 6
At C (4, 0) ⇒ Z = -12
At D (0,0) ⇒ z = 0
Hence Z is minimum at C (4, 0) = -12
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.2

KSEEB Solutions

Question 3.
Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15,5x +2y ≤ 10, x ≥ 0,y ≥ 0
Answer:
3x + 5y ≤ 15

X05
y30

0 + 0 <15
true, toward 0
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.3

5x +2y = 10

X20
y05

ABCD is the solution region
At A(0,3) ,Z = 3 x 3 = 9

Question 4.
Minimum Z = 3x + 5y
such that x +3y ≤ 3, x+y ≥ 2, x, y  ≥0.
Answer:
x + 3y = 3

X30
y01

0 + 0,3 false
∴ shading away from 0
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.4
x + y = 2

X021
y201

0 + 0 ,2 false
shading away from 0
Away from 0
corner points of feasible region are
A (0,2); Z = 5 x 2 = 10
B (1.5, 0.5); Z = 7
C (3,0); Z = 3 x 3 = 9
∴ Z is minimum at B (1.5,0.5) = 7.

KSEEB Solutions

Question 5.
Maximise Z = 3x +2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Answer:
x + 2y = 10

X064
y543

0 + 0 < 10
True
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.5

3y + y = 15

X543
y036

0 + 0 < 15
True

ABCO is the solution region
At A (0,5) Z = 2 x 5 = 10
At B (4, 3) Z = 3 x 4+ 2 x 3 = 18
At C (5,0) Z = 3 x 5 = 15
∴ Z is maximized at B (4, 3) = 18.

Question 6.
Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0
Answer:
2x + y = 3

X10
y13

0 + 0, 0 false
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.6

x + 2y = 6

X602
y032

0 + 0 ,6 false
An and above O is the solution region
At A(0,3) Z = 2 x 3 = 6
At B(6,0) Z = 6
Z is maximum at all points on line AB = 6
Here minimum of Z occurs at 2 points A & B.

KSEEB Solutions

Show that the minimum of Z occurs at more than two points.

Question 7.
Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0,x,y ≥ 0.
Answer:
x + 2y = 120

X60020
Y306050

0 + 0 ,120
True

x + y = 60

X302040
Y304020

0 + 0 >0
false
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.7

x – 2y =0

x20040
y10020

40 – 20 > 0
ABCD is the solution region
At A (60, 30) Z = 5 x 60 + 10 x 30 = 600
At B (40, 20) Z = 5 x 40+10 x 20 = 400
At C (60,0) Z = 5 x 60 = 300
At D (120,0) Z = 5 x 120 = 600
Z is minimized at C (60,0) = 300
Z is maximized at 2 points A (60,30), D (120,0)
∴ maximum value is 600 at all
points on the line joining A (60,30) and D (120,0)

KSEEB Solutions

Question 8.
Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
Answer:
x + 2y = 100

x1005020
y02040

0 + 0>100
0 – 0 < 0 False

2x – y = 0

x1002010
y2004020

0 – 0 < 0
False

2x +y = 200

x1005060
y010080

0 + 0 <200
True
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.8

ABCD is the solution                              .
At A (0,200),          Z = 2 x 200 = 400
At B (50,100), Z = 50 + 2 x 100 = 250
At C (20,40), Z = 20 + 2x 40= 100
At D (0,50), Z = 2 x 50 =100
Z is maximised at A (0,200) = 400
Z is minimised at 2 points C (20,40) & D (0, 50) = 100

KSEEB Solutions

Question 9.
Maximise Z = -x + 2y, subject to the constraints:
x ≥ 3, x + y ≥ 5 , x + 2y ≥ 6, y ≥ 0
Answer:
x + y = 5

x132
y423

0 + 0, 5
False

0,3
False
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.9

x + 2y = 6

x042
y312

0 + 0
False

ABC is the solution region
At A (3,2)       Z =-3+2 x 2=1
At B (4,1)       Z = -4 + 2 x 1 = -2
At C (6, 0)      Z = -6 = -6
Z is maximised at A (3,2) = 1 and
minimised at C (6,0) = -6

KSEEB Solutions

Question 10.
Maximise Z = x + y,
subject to x – y ≤ -1, -x +y ≤ 0, x, y ≥ 0.
Answer:
x – y = -1

x102
y213

0 – 0 ,-1
False

x + 2y = 6

x042
y312

0 + 0 , 0 False,
Since there is no common shaded region, there is no feasible solution region. Hence maximum of Z does not exist.
2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1.10

a